First of all, if you’re taking the November test, good luck. If you’re prepared well, you have nothing to fear. Get in there and beast it.

Of course, many people aren’t taking the November SAT, and they deserve some love, too. If you’re not taking the test tomorrow, here’s one tough question to noodle on while your contemporaries are out achieving glory.
Prize for the first correct response: access to the PWN the SAT Math Guide. You must not be anonymous to win.
In the figure above, B is the midpoint of AC and the center of the green circle. All other labeled points are also the centers of circles. If the green area is 9π, what is the area of the circle with center E?

I actually cackled sinisterly when I made this diagram. Good luck! 🙂

UPDATE: Congrats to “emolano82” for getting it first. Solution below!

There’s no “shortcut” here, other than to rub your hands together and say “OH YEAH!” like Macho Man Randy Savage. If you’re not afraid of a bunch of circles, you’ll be OK.

The key here is to take it one circle at a time. Start with the green one. If its area is 9π, then it has to have a radius of 3, right? And a diameter of 6. Which is important because…

It’s the radius of the red circle. So…so far we know BC = 3, and that the radius of the red circle above is 6. So EC = 6, because E is on the circle, and C is its center.

Looky looky! That’s a right triangle. The short leg is 3, and the hypotenuse is 6. So you can pythagorize, or you can remember your special rights and know that this is a 30°-60°-90° right off the bat, so the long leg must be 3√3. EB = 3√3.

You can do the same thing below; BD = 3√3. So ED, which is the radius of the circle with center E, must be 6√3.

To find the area:

Area = π(6√3)2
Area = 108π Emolano82 says:

108 pi. Anthony B. says:

81pi? Stella Rachelle says:

144 pi? Cindy Nguyen says:

144 pi..? Doc G says:

I’m gettin’ 108pi.

The key is to realize that you’re looking for a relationship between ED and AC. You get it because AE and AC are both the same length, since E and C are both on circle A (and of course the area of the green circle gives you AB = AC/2).
If you actually draw in the line AE, you see that it could be the hypotenuse of a right triangle with AB and EB as legs, which is useful because you know AB. So EB^2 = AE^2 – EB^2.

Similar reasoning then gives BD = EB, so ED, the radius of the large circle, is 2EB and you can calculate the area from there:
Green area = 9pi -> AB = 3
AC = 2AB = 6 = AE
EB = sqrt(36 – 9) = sqrt(27)
ED = 2*sqrt(27) -> area of circle E = pi*ED^2 = pi*4*27 = 108pi

Now… what’s the smart SAT trick way to do it? Mike McClenathan says:

No tricks, just treats. (I don’t know what that means.) Mathemagician1729 says:

Hello yes the answer is 108pi

to see this simply extend the inradius of 3 for the green circle until the entire chord spanning circle E is covered. The length for this is 18. Now you see by symmetry that you have a right triangle with legs r/2, 9 and hypotenuse r. An application of pythagorean principle immediately reveals the answer. Dodo1995 says:

I got 108 pi as well. Akil Bello says:

you are an evil genius!